In ΔABC shown below, segment DE is parallel to segment The following two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally:

Statement Reason
1. Line segment DE is parallel to line segment AC 1. Given
2. Line segment AB is a transversal that intersects two parallel lines. 2. Conclusion from Statement 1.
3. ∠BDE ≅ ∠BAC 3. Corresponding Angles Postulate
4. 4.
5. 5.
6. BD over BA equals BE over BC 6. Converse of the Side-Side-Side Similarity Theorem

Which statement and reason accurately completes the proof?

4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate
5. ∠B ≅ ∠B; Reflexive Property of Equality
4. ∠B ≅ ∠B; Reflexive Property of Equality
5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate
4. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate
5. ∠A ≅ ∠C; Isosceles Triangle Theorem
4. ∠A ≅ ∠C; Isosceles Triangle Theorem
5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate